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ACID-BASE EQUILIBRIA Definition of acids and bases – Arrhenius – Brønsted-Lowry Strong vs weak acids and bases Self-ionization of water (K w ) Ionisation.

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Presentation on theme: "ACID-BASE EQUILIBRIA Definition of acids and bases – Arrhenius – Brønsted-Lowry Strong vs weak acids and bases Self-ionization of water (K w ) Ionisation."— Presentation transcript:

1 ACID-BASE EQUILIBRIA Definition of acids and bases – Arrhenius – Brønsted-Lowry Strong vs weak acids and bases Self-ionization of water (K w ) Ionisation constants (K a, K b ) pH, pOH

2 ARRHENIUS An Arrhenius acid is a substance that, when dissolved in water, increases the concentration of hydronium ions, H 3 O + (aq). An Arrhenius base is a substance that, when dissolved in water, increases the concentration of hydroxide ions, OH ¯ (aq).

3 BRØNSTED-LOWRY Acid: proton donor Base: proton acceptor Conjugate acid/base pairs: The stronger the acid, the weaker is its conjugate base, and vice versa.

4 HF + H 2 O ⇌ H 3 O + + F ¯ acidbase acid base conjugate pair

5 NH 3 + H 2 O ⇌ NH 4 + + OH ¯ baseacid acid base conjugate pair

6 Any species that is an Arrhenius acid or base is also a Brønsted-Lowry acid or base. Brønsted-Lowry theory accounts for amphoteric species (e.g. H 2 O). *Amphoteric: can act as acid or base.

7 PRACTICE EXAMPLES (a)HPO 4 2- + NH 4 + ⇌ NH 3 + H 2 PO 4 ¯ (b)SO 4 2- + H 2 O ⇌ HSO 4 ¯ + OH ¯

8 Strong and weak acids Strong acids are completely ionised in water. e.g. HCl  H 3 O + + Cl ¯ Weak acids are partially ionised in water. e.g. HF + H 2 O ⇌ H 3 O + + F ¯ H2OH2O

9 Strong and weak bases Strong bases are completely ionised in water. Have OH ¯ in their formula. e.g. NaOH  Na + + OH ¯ Weak bases are partially ionised in water. Typically contain nitrogen. e.g. NH 3 + H 2 O ⇌ NH 4 + + OH - H2OH2O

10 STRONG ACIDSSTRONG BASES HClLiOH HBrNaOH HIKOH HClO 3 RbOH HClO 4 CsOH HNO 3 Ca(OH) 2 H 2 SO 4 Sr(OH) 2 Ba(OH) 2

11 WEAK ACIDSWEAK BASES CH 3 COOHNH 3 HClO 2 C 6 H 5 NH 2 HCOOHHONH 2 HNO 2 CH 3 NH 2 HOClC5H5NC5H5N HF HCN

12 SELF-IONIZATION OF WATER H 2 O (l) ⇌ H + (aq) + OH ¯ ( aq ) Strictly speaking the hydrogen ion H + exists in water as the hydronium ion, H 3 O +. The dissociation of water should therefore really be written as 2 H 2 O (l) ⇌ H 3 O + (aq) + OH ¯ ( aq )

13 Since the water is only slightly ionized, the molar concentration of water has essentially a constant value of 55.6 M.  [H 2 O] 2 is a constant which we combine with K c. [H + ][OH  ] = K c  [H 2 O] 2  [H + ][OH ¯ ] = constant constant = K w K w is the ionic product of water = 1.00 x 10  14

14 IONIZATION CONSTANTS Large values for strong acids and bases. Small values for weak acids and bases. ACID: K a HA (aq) + H 2 O (l) ⇌ H 3 O + (aq) + A ¯ (aq) *H 2 O (l) excluded since it is a liquid

15 BASE: K b B (aq) + H 2 O (l) ⇌ BH + (aq) + OH¯ (aq) *H 2 O (l) excluded since it is a liquid

16 The product of the ionization constant for an acid and the ionization constant of its conjugate base is the ionic product of water. K a x K b = K w

17 Hydrofluoric acid HF (aq) + H 2 O (l) ⇌ H 3 O + (aq) + F ¯ (aq) Fluoride ion (its conjugate base) F ¯ (aq) + H 2 O (l) ⇌ HF (aq) + OH ¯ (aq) = [H 3 O + ][OH ¯ ] = K w

18 pH and pOH pH = -log[H + ] pOH = -log[OH ¯ ] K w = [H + ][OH ¯ ] log K w = log[H + ] + log[OH ¯ ]  log K w =  log[H + ] + (  log[OH ¯ ]) pK w = pH + pOH

19 In pure water, [H + ] = [OH ¯ ] = = 10  7 M A solution in which [H + ] = [OH ¯ ] is a neutral solution. If [H + ] > 10  7 M the solution is acidic, and if [H + ] < 10  7 M the solution is alkaline (or basic).

20 If [OH ¯ ] > 10  7 M the solution is alkaline; if [OH ¯ ] < 10  7 M the solution is acidic. Hence measurement of either [H + ] or [OH ¯ ] yields the other. or

21 At 25  C pK w = pH + pOH = 14 neutral solution:pH = pOH = 7 acidic solution:pH 7 alkaline solution:pH > 7 pOH < 7

22 ICE TABLES FOR EQUILIBRIUM CALCULATIONS ABCD INITIAL[A] 0 [B] 0 00 CHANGES-ax-bx+cx+dx EQUILIBRIUM[A] 0 - ax[B] 0 - bxcxdx aA + bB ⇌ cC + dD ⇌

23 PRACTICE EXAMPLE Calculate the pH of a 0.050 M nitrous acid (HNO 2 ) solution. K a = 4.5 x 10 -4

24 PRACTICE EXAMPLE What is the pH of a 0.10 M solution of NH 3 ? What is the % ionisation for this solution? K b = 1.8 x 10  5


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